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Find different possible combinations of the rows of two matrices.

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Consider two matrices:
a = [1 2 3; 4 5 6; 7 8 10]
a =
1 2 3
4 5 6
7 8 10
and b=a + 10
b =
11 12 13
14 15 16
17 18 20
first, I am trying to obtain all possible combination of rows (1st row of 'a' and all rows of 'b', second row of 'a' and all rows of 'b' and so on) such that the output is
c =
1 2 3 11 12 13
4 5 6 11 12 13
7 8 10 11 12 13
1 2 3 14 15 16
4 5 6 14 15 16
7 8 10 14 15 16
1 2 3 17 18 20
4 5 6 17 18 20
7 8 10 17 18 20
secondly for each row, I am storing the first 3 and the last 3 columns in a single index of a 3 index vector. So in this example the vector of combosets is 2x3x9 where element is
combosets(:,:,1) =
1 2 3
11 12 13
combosets(:,:,2) =
4 5 6
11 12 13
and so on. The code I am using is below. Is there any better and faster way to do this and prefarably step 1 and 2 combined?
Thanks in advance!
clc
clearvars;
a = [1 2 3; 4 5 6; 7 8 10]; % 3 by 3 matrix
b=a + 10; % 3 by 3 matrix
r=size(a,1);
combosets=zeros(r-1,r,r*r);
d={NaN};
% Step 1: storing different combination of rows in cell
for k=1:1:r
c=vertcat(horzcat(a(:,:),repmat(b(k,:),r,1)),horzcat(a(:,:),repmat(b(k,:),r,1)),horzcat(a(:,:),repmat(b(k,:),r,1)));
d{k}=(unique(c,'rows'));
end
ddash=cell2mat(d'); % making it a matrix again
% Step 2: storing in a multidimensional array
for j=1:1:r*r
rowset1=ddash(j,1:3);
rowset2=ddash(j,4:6);
combosets(:,:,j)=[rowset1;rowset2];
end

Accepted Answer

Johannes Fischer
Johannes Fischer on 8 Sep 2020
a = [1 2 3; 4 5 6; 7 8 10]; % 3 by 3 matrix
b=a + 10; % 3 by 3 matrix
r=size(a, 1);
% create repetitions of values
a = repmat(a, [r, 1]);
b = repelem(b, r, 1);
% create combination matrix
c = [a b];
% reshape into combosets
combosets = reshape(c', 3, 2, 9);
% permute to get dimensions right
combosets = permute(combosets, [2, 1, 3]);
  1 Comment
Tuhin Choudhury
Tuhin Choudhury on 8 Sep 2020
Thank you! works perfectly fine. Infact your method works fastest and so I accepted this answer . although all of them work perfectly fine

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More Answers (1)

Adam Danz
Adam Danz on 8 Sep 2020
Edited: Adam Danz on 8 Sep 2020
Fast & clean 2-liner:
a = [1 2 3; 4 5 6; 7 8 10];
b = a + 10;
c = [repmat(a,size(b,1),1), repelem(b, size(a,1),1)];
combosets = permute(reshape(c.',3,2,9),[2,1,3]);
  5 Comments
Adam Danz
Adam Danz on 8 Sep 2020
"I wish I could accept more than 1 answer"
You accepted the right one if it works for your data and it make sense to you. The time difference is so small that readability and understandability are important factors. Plus, I like when newer contributors get recognition for their contributions and encouragement to continue doing so 🙂.

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